Uttarakhand Board Class 12 Sample Paper 2026 Maths

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UBSE Sample Question Papers 2026

a jksy ua /Roll No.
eqfnr i`’Bksa dh la[;k /No of
printed pages: 7

128 428(---)

Ikzfrn”kZ iz”ui= 2026 / Model Paper 2026
xf.kr MATHEMATICS
le; 3 ?kaVs iw.kkaZd 80
Time : 3 Hours Max. Marks: 80

funsZ'k (i)bl ç'u i= esa dqy 24 ç'u gS lHkh ç'u vfuok;Z gSa
Directions : There are 24 questions in the question paper. All questions are Compulsory.

(ii) ç'u gsrq fu/kkZfjr vad muds lEeq[k vafdr gSAa
Marks allotted to the questions are mentioned against them.

(iii) çR;sd ç'u dks /;ku iwoZd if<+, rFkk leqfpr mÙkj nhft,A
Read each question carefully and answer to the point.

(iv) ç'u la[;k 1 cgqfodYih; ç'u gS bl ç'u ds çR;sd [kaM ds mÙkj esa pkj fodYi fn, x, gSa
lgh fodYi viuh mÙkj iqfLrdk esa fyf[k,A
Question no. 1 is a multiple choice question. Four options are given in answer to part of this question.
Write the correct option in your answer book.

(v) ç'u la[;k 1 dk çR;sd [kaM ,d vad dk gSA ç'u la[;k 2 ls 7 rd ,d vad ds ç'u gSaA
ç'u la[;k 8 ls 12 rd nks vad ds ç'u gSaA ç'u la[;k 13 ls 18 rd pkj vad ds ç'u gSaA ç'u
la[;k 19 ls 24 rd 5 vad ds ç'u gSa] ftuesa ls ç'u la[;k 24 dsl lzksr vk/kkfjr ç'u gSA
Each part of question no. 1 carries one mark. Question no. 2 to 7 are of one mark each. Question no. 8
to 12 are of two marks each. Question no. 13 to 18 are of four marks each. Question no. 19 to 24 are
of five marks each, in which question no. 24 is a Case/Source based question.

(vi) bl ç'u i= esa lexz ij dksbZ fodYi ugha gS rFkkfi dfri; ç'uksa esa vkarfjd
fodYi çnku fd;k x;k gSA ,sls ç'uksa esa dsoy ,d fodYi dk gh mÙkj nhft,A
There is no overall choice in this question paper, however, an internal choice has been provided in a
few questions. Attempt only one of the given choices in such questions.

1¼d½ eku yhft, fd Qyu f:R→R f(x) =x2 }kjk ifjHkkf"kr gS] lgh mÙkj dk p;u dhft, (1)
(i) f ,dSdh vkPNknd gS
(ii) f ,dSdh gS vkPNknd ugha
(iii) f vkPNknd gS ,dSdh ugha
(iv) fa u rks ,dSdh gS u gh vkPNknd
Let f:R→R is defined as f(x) =x2. Choose the correct answer:
(i) f is one-one onto
(ii) f is one-one but not onto
(iii) f is onto but not one-one
(iv) f is neither one-one nor onto

¼[k½ vkO;wg A= [aij]m×n , aij =0 ;fn i ≠j dgykrk gS (1)
(i) fod.kZ vkO;wg
(ii) vfn'k vkO;wg
(iii) rRled vkO;wg
(iv) 'kwU; vkO;wg
Matrix A= [aij]m×n , aij =0 when i ≠j is known as :

(i) Diagonal matrix
(ii) Scalar matrix
(iii) Identity matrix
(iv) Zero matrix

¼x½ ;fn If x-y= rks cjkcj gS : (1)

If x-y= then is equal to :

(i) 0 (ii) 1 (iii) -1 (iv)

¼?k½ ;fn ,d f(x)le Qyu gS rks ∫ 𝑓(𝑥)𝑑𝑥 cjkcj gS (1)
If f(x) is an even function then∫ 𝑓(𝑥)𝑑𝑥 is equal to :

(i) 0 (ii) π/2 (iii) 2∫ 𝑓(𝑥)𝑑𝑥 (iv) ∫ 𝑓(𝑥)𝑑𝑥
(1)

2
¼M+½ tan dk çfr&vodyt cjkcj gS
Anti derivative of tan2 is equal to :

(i) tan -x+c

(ii) tan -x +c
(iii) sec4 +c
(iv) 2 sec4 +c
¼p½ vody lehdj.k y23 +y2 +sin y1 =0 dh dksfV ,oa ?kkr Øe'k% gS (1)
(i) 3,2 (ii) vifjHkkf’kr ,2 (iii) 3, vifjHkkf’kr (iv) 1,2

The order and degree of the differential equation y23 +y2 +sin y1 =0 are respectively:
(i) 3, 2 (ii) Not defined, 2 (iii) 3, Not defined (iv) 1, 2

¼N½ y v{k ds fnd dkslkbu gSa (1)
Direction cosine of y-axis is :
(i) (1,0,0) (ii) (1,0,1) (iii) (0,1,0) (iv) (1,1,1)

¼t½;fn A vkSj B nks ?kVuk,a bl çdkj gSa fd P ( ) = P( ) rc (1)

If A and B are two events such that P ( ) = P( ) then:

(i) P(A) =0 (ii) P(B) =0 (iii) P(A⋂B) =0 (iv) P(A)-P(B) =0

funsZ'k ç'u la[;k 1 ds vxys nks [k.Mksa esa nks dFkuksa dks vfHkdFku (A) rFkk dkj.k ( R ) ds :i
esa fpfUgr fd;k x;k gSA fuEufyf[kr fodYiksa (i) , (ii) , (iii) ,and (iv) esa ls pqudj budk lgh mÙkj
nhft,
Reason ( R ) . From the following options (i) , (ii) , (iii) ,and (iv), select their correct answer .
(i) A vkSj B nksuksa lgh gSa vkSj A, B dh lgh O;k[;k djrk gSA
Both A and B are correct and R is the correct explanation of A

(ii) A vkSj B nksuksa lgh gSa ijarq A, B dh lgh O;k[;k ugha djrk gSA
Both A and B are correct but R is not the correct explanation of A

(iii) A lgh gS ijarq R xyr gSA
A is correct but R is incorrect

(iv) A vkSj B nksuksa xyr gSA
a
Both A and B are incorrect

>) vfHkdFku (A) : f(x) =|cos x| ,d larr Qyu gS
Assertion (A) : f(x) = Acos x| is a continuous function. (1)

dkj.k ( R ) g(x) = cos x vkSj h(x)=|x| larr Qyu gSaA
Reason (R) : g(x) =cos x and h(x)=|x| are continuous functions.

¥) vfHkdFku (A) : dksbZ Hkh nks lfn'kksa 𝑎⃗ vkSj 𝑏⃗ ds fy, 𝑎⃗. 𝑏⃗ ≤ |𝑎⃗| 𝑏⃗

Assertion (A) : For any two vectors 𝑎⃗ and 𝑏⃗ , 𝑎⃗. 𝑏⃗ ≤ |𝑎⃗| 𝑏⃗ (1)

dkj.k Reason (R) : cos 𝜃 ≤1

2. ;fn A = { 1,2,3 }, rks Als A esa lac/a kksa dh la[;k dks Kkr dhft,
If A = { 1,2,3 }, find the number of relations from A to A. (1)

3. ;fn x = a 𝜃2 , y = 2 a 𝜃 rks dk eku Kkr dhft,

If x = a 𝜃2 , y = 2 a 𝜃 find : (1)

4. eku Kkr dhft, Evaluate

√
∫ dx (1)
√

5. vody lehdj.k dk lekdyu xq.kkad Kkr dhft,

Find the integrating factor of the differential equation (1)

y1 + y sec x =tan x ( 0 ≤ x ≤ π/2 )

6. lfn'k 𝚤̂ + 𝚥̂+ 𝑘 ,oa x- v{k ds chp dk dks.k Kkr djksA (1)
Find the angle between x- axis and the vector i + j + k .

7. fuEufyf[kr js[kk dk lehdj.k dkrhZ; :i esa fy[kks A (1)
Write the equation of the line in cartesian form

r = (5i -4 j +6 k ) + λ(3i + 7j + 2k )
√ √
8 ljy dhft, @Simplify cos -1 ( ) + sin -1 ( ) (2)
3
9. fdlh ?ku dk vk;ru 7 lseh @lsd.M dh nj ls c< jgk gS ] mldk i`’B {ks=Qy fdl nj ls
c< jgk gS tc mldh dksj 4 lseh gS A (2)

The volume of a cube is increasing at a rate of 7 cm3 / sec. How far is the surface area increasing
when the length of an edge is 4 cm?

10. eku Kkr dhft, Evaluate ∫{ }dx . (2)

11. fl} dhft, fd fcanq A(2,3,-4) , B(1,-2,3) vkSj C(3,8,-11) laj[
s k gSaA (2)

Prove that the points A(2,3,-4) , B(1,-2,3) and C(3,8,-11) are collinear .

12.A vkSj B nks [kkyh inksa ds fy, mEehnokj gSaA ;fn A ds pquus dh çkf;drk vkSj B ds pqus
tkus dh çkf;drk gS rks Bhd Bhd ,d ds pqus tkus dh çkf;drk Kkr djksA (2)

A and B are candidates for two vacant posts. If the probability of A being selected is and the

probability of B being selected is ,find the probability that exactly one of them is selected .

13. eku yhft, leLr n ∊ N ds fy, f(n) = , ;fn n fo"ke gS

, ;fn n le gS

}kjk ifjHkkf’kr ,d Qyu f: N→ R gS
Kkr dhft, fd D;k Qyu f ,dSdh vkPNknh gSA (4)

If f: N→ R be defined by f(n) = , if n is odd
for all n ∊ N

, if n is even ,

Find whether the function is bijective ?
Or

;fn L ry esa lHkh js[kkvksa dk leqPp; gS vkSj R = { (L1,L2) : L1 yacor gS L2 ij } leqPp; esa ,d
ifjHkkf"kr laca/k gSA tkafp, fd R lHkh laca/kksa ds çdkj esa fdl çdkj dk laca/k gS
If L is a set of all lines of the plane and R = { (L1,L2) : L1 is perpendicular to L2 } is a defined
relation in set L , examining the types of relations , tell what types of relations is R ?

0 𝑎 3
14 ;fn vkO;wg A= 2 𝑏 −1 ,d fo’ke lefer vkO;wg gS rks a, b ,oa c ds eku Kkr dhft,A
𝑐 1 0

0 𝑎 3
If matrix A= 2 𝑏 −1 is a skew-symmetric matrix then finds the values of a, b and c. (4)
𝑐 1 0

15.;fn y = (x)x + (log x)x , Kkr dhft,A (4)

If y = (x)x + (log x)x , find

Or

k dk eku Kkr dhft, ;fn Qyu

f(x) = , if x ≠ 0

, if x =0 , x =0
K
ij larr gS

Find the value of k , if the function
2
f(x) = (1-cos 4x)/8 x , if x ≠ 0

K , if x =0 is continuous at x =0 .

16.
eku Kkr dhft, / Evaluate (4)
√( )
∫√( )
dx

or

( )( )
dk x ds lkis{k lekdyu Kkr dhft,A

Find the integration of ( )( )
with respect to x.

17. ml f=Hkqt dk {ks=Qy Kkr dhft, ftlds 'kh"kZ A(1,-1,2) , B(2,1,-1) vkSj C(3,-1,2) gSaA (4)

Find the area of a triangle whose vertices are A(1,-1,2) , B(2,1,-1) and C(3,-1,2).

18. nh xbZ js[kkvksa ds chp dh U;wure nwjh Kkr djksA (4)

Find the shortest distance between the lines given by
r = (8+3λ)i - (9+16λ) j +(10+7λ) k and
r = 15i + 29 j +5 k + (3i + 8 j -5 k)

2 7 6 9
19. ;fn A= vkSj B=
3 5 7 8
rks tkafp, D;k (AB)-1 = B-1A-1 (5)

If A= 2 7 and B= 6 9
3 5 7 8
then verify that (AB)-1 = B-1A-1
or

fuEufyf[kr lehdj.k fudk; dks vkO;wg fof/k ls gy dhft,A
Solve the following system of equations by using matrix method :

+ + =6

- + =5

+ - =7

20. fl} dhft, fd ,d lef}ckgq f=Hkqt dk U;wure ifjeki ftlesa f=T;k r dk ,d o`Rr vafdr
fd;k tk ldrk gS ] 6√3 r gS A (5)

Prove that the least perimeter of an isosceles triangle in which a circle of radius r can be inscribed is
6√3 r.
Or

fl) dhft, fd fdlh xksys ds varxZr vf/kdre vk;ru okyk 'kadq ,oa xksys ds vk;ruksa esa 8%27
dk vuqikr gSA
Prove that the ratio of volume of the maximum cone inscribed in a sphere of radius R is 8:27 to the
volume of the sphere.

21. js[kk x= a/2 }kjk o`Ùk x2 + y2 = a2 ds dkVs x, NksVs Hkkx dk {ks=Qy gS (5)
2 2
Find the area of the smallest part of the circle x + y = a2, cut off by the line x= a/2.
22. ml oØ dk lehdj.k Kkr dhft, tks ewy fcanq ls tkrk gS ;fn dy/dx =(x+y) /(x-y) .

Find the curve which passes through the origin if dy/dx =(x+y) /(x-y) . (5)
Or

vody lehdj.k (tan -1x- y) dx = (1+x2) dy dks gy djks
Solve the differential equation (tan -1x- y) dx = (1+x2) dy

23. vkys[kh; fof/k ls fuEufyf[kr O;ojks/kksa ds varxZr dk vf/krdehdj.k Kkr dhft,A (5)

Determine graphically the maximum value of the given objective function subject to constraints :

Mn~n”s ; Qyu / Objective function : z =1000x + 600y

O;ojks/k / Constraints : x + y ≤ 200

4x + y ≤ 0
x ≥ 20 , x ≥ 0 , y ≥ 0

24. Nk= i<+kbZ ds ?kaVks ds vk/kkj ij fuEu çdkj ls oxhZ—r fd, x, gSa

Xkzqi A ( 2 ?kaVs ls de) : 30%
Xkzqi B (2 ls 4 ?kaVs) :50%
Xkzqi C (4 ?kaVs ls vf/kd) : 20%
;fn çR;sd xzqi esa 80% ls vf/kd çkIr djus okys Nk= Øe'k% 20% ,50% vkSj 70% gSa

;fn d{kk ls ,d Nk= dks ;k–f'pd pquk tkrk gS rks çkf;drk Kkr djsa
(i) Nk= 4 ?kaVs çfrfnu i<+kbZ ds lewg ls pquk tkrk gS vkSj 80% vad çkIr djrk gS A (1)
(ii) Nk= 80% vad çkIr djrk gS A (2)

(iii) Nk= 2 ?kaVs çfrfnu ds lewg ls pquk tkrk gS ;fn ;g Kkr gS fd mlus 80% vad
çkIr fd, gSa A (2)

Students were divided , according to their study hours :
Group A (Less than 2 hours ) : 30%
Group B ( 2 to 4 hours ) : 50%
Group A (More than 4 hours ) : 20%
The % of students scoring above 80% in each group was 20% , 50% and 70% respectively.
If a student is selected at random from the class, find the probability that:
(i)The student studies more than 4 hours per day and scores above 80%.

(ii) The student scores above 80%.

(iii) The student studies less than 2 hours per day given that he/she scored 80%.

a jksy ua /Roll No. eqfnr i`’Bksa dh la[;k /No of printed pages: 7

128 428(---)

Ikzfrn”kZ iz”ui= 2026 / Model Paper 2026
xf.kr MATHEMATICS
le; 3 ?kaVs iw.kkaZd 80
Time : 3 Hours Max. Marks: 80

funsZ'k (i)bl ç'u i= esa dqy 24 ç'u gS lHkh ç'u vfuok;Z gSa
Directions : There are 24 questions in the question paper. All questions are Compulsory.

(ii) ç'u gsrq fu/kkZfjr vad muds lEeq[k vafdr gSaA
Marks allotted to the questions are mentioned against them.

(iii) çR;sd ç'u dks /;ku iwoZd if<+, rFkk leqfpr mÙkj nhft,A
Read each question carefully and answer to the point.

(iv) ç'u la[;k 1 cgqfodYih; ç'u gS bl ç'u ds çR;sd [kaM ds mÙkj esa pkj fodYi fn, x, gSa
lgh fodYi viuh mÙkj iqfLrdk esa fyf[k,A
Question no. 1 is a multiple choice question. Four options are given in answer to part of this question.
Write the correct option in your answer book.

(v) ç'u la[;k 1 dk çR;sd [kaM ,d vad dk gSA ç'u la[;k 2 ls 7 rd ,d vad ds ç'u gSaA
ç'u la[;k 8 ls 12 rd nks vad ds ç'u gSaA ç'u la[;k 13 ls 18 rd pkj vad ds ç'u gSaA ç'u
la[;k 19 ls 24 rd 5 vad ds ç 'u gSa] ftuesa ls ç'u la[;k 24 dsl lzksr vk/kkfjr ç'u gSA
Each part of question no. 1 carries one mark. Question no. 2 to 7 are of one mark each. Question no. 8
to 12 are of two marks each. Question no. 13 to 18 are of four marks each. Question no. 19 to 24 are
of five marks each, in which question no. 24 is a Case/Source based question.

(vi) bl ç'u i= esa lexz ij dksbZ fodYi ugha gS rFkkfi dfri; ç'uksa esa vkarfjd
fodYi çnku fd;k x;k gSA ,sls ç'uksa esa dsoy ,d fodYi dk gh mÙkj nhft,A
There is no overall choice in this question paper, however, an internal choice has been provided in a
few questions. Attempt only one of the given choices in such questions.

1 ( d) rhu Nk=ksa vkHkk ]Hkfor vkSj fpjkx ls X= ø1]3]5]7]9À leqPp; dks
Y= ø2]4]6]8À leqPp; ds fy;s ,d Qyu ifjHkkf’kr djus dks dgk x;kAmuds mRrj uhps
fn[kk, x, gSaA vkHkk % f={(1,2), (1,4), (1.6), (1,8) }, Hkfor %f={(1,2), (3,4), (5.6), (7,8) }fpjkx
f={(1,4), (3,4), (5.4), (7,4), (9,4) }
fdlus lgh rjg ls Qyu ifjHkkf’kr fd;k ? (1)
(i) dsoy fpjkx
(ii) dsoy Hkfor vkSj fpjkx
(iii) dsoy vkHkk vkSj Hkfor
(iv) dsoy Hkfor
Three students Abha ,Bhavit and Chirag were asked to define a function X= ø1]3]5]7]9À to set
.There responces are shown below,
Abha : f={(1,2), (1,4), (1.6), (1,8) } , Bhavit : f={(1,2), (3,4), (5.6), (7,8) } ,
Chirag : f={(1,4), (3,4), (5.4), (7,4), (9,4) }
Who defined a function correctly ?
(i) Only Chirag
(ii) Only Bhavit and Chirag
(iii) Only Abha and Bhavit
(iv) Only Bhavit
√
1 ( [k) ;fn Sin( sin -1 ( ) +cos -1𝑥 )=1 rc x dk eku gS& (1)
-1 √
If Sin( sin ( ) +cos -1 𝑥 )=1 then x is equal to
√
(i) 1 (ii) 0 (iii) 4/5 (iv)
5 𝑥
¼x½ ;fn A= 𝑦 0 vkSj A= AT rc (1)

5 𝑥
If A=
𝑦 0
and A= AT then

(i) x=0 ,y=5 (ii) x+y=5 (iii) x=y (iv) none of these
3
¼?k½ x log x dk x ds lkis{k f}rh; dksfV dk vodyt gS& (1)
The second order derivative of x3log x w.r.to x.
(i) x(5+6 log x) (ii) x3 (5+6 log x) (iii) x(6+5 log x) (iv) x2 (6+5 log x)

¼M+½ ∫ dx dk eku gksxk (1)

∫ dx is equal to

(i) 2 Sin x + log (Sec x –tan x) + C (ii) 2 Sin x - log (Sec x –tan x) + C
(iii) 2 Sin x + log (Sec x +tan x) + C (iv) -2 Sin x - log (Sec x –tan x) + C
2 2
vody lehdj.k d y2 + x dy = 2 x2 log d y2 dh dksfV gS&
2
¼p+½ (1)
dx

dx dx
(i) 2 (ii) 1 (iii) 3 (iv) vifjHkkf’kr

The order of differential equation +x ( ) = 2 x2 log ( ) is

(i) 2 (ii) 1 (iii) 3 (iv) Undefined

¼N+½ fcUnq (2,5,7) ls x v{k dh nwjh gS& (1)

The distance of X axis from the point (2,5,7)

(i) 2 (ii) √74 (iii) 29 (iv) 53

¼t+½ nks xsanas ,d ckDl esa ls fcuk izfrLFkkfir fd, fudkyh tkrh gSaAckDl esa 10 dkyh vkSj 8
yky xsna s gSa rks izFke ds dkyh ,oa nwljh ds yky gksus dh izkf;drk gS& (1)

Two balls are drawn without replacement from a box. The box contains 10 black and 8 red
balls. The probability that the first is black and the second is red is

(i) (ii) (iii) (iv) buesa ls dksbZ ugha
funsZ'k ç'u la[;k 1 ds vxys nks [k.Mksa esa nks dFkuksa dks vfHkdFku (A) rFkk dkj.k ( R ) ds :i
esa fpfUgr fd;k x;k gSA fuEufyf[kr fodYiksa (i) , (ii) , (iii) ,and (iv) esa ls pqudj budk lgh mÙkj
nhft,
Reason ( R ) . From the following options (i) , (ii) , (iii) ,and (iv), select their correct answer .
(i) A vkSj B nksuksa lgh gSa vkSj A, B dh lgh O;k[;k djrk gSA
Both A and B are correct and R is the correct explanation of A

(ii) A vkSj B nksuksa lgh gSa ijarq A, B dh lgh O;k[;k ugha djrk gSA
Both A and B are correct but R is not the correct explanation of A

(iii) A lgh gS ijarq R xyr gSA
A is correct but R is incorrect

(iv) A vkSj B nksuksa xyr gSaA
Both A and B are incorrect

>) vfHkdFku (A) : |Sin x| izR;sd x ∈ R ds fy, ,d larr Qyu gS
Assertion (A) : |Sin x| is a continuous function for all x∈ R (1)

dkj.k ( R ) Sinx vkSj |x| , R esa larr Qyu gSaA
Reason (R) : Sinx and |x| , are continuous in R .

¥) vfHkdFku (A) : lfn”k 𝑎⃗ = 2i - j +5 k vkSj 𝑏⃗ =5i + j -3 k ,d nwljs ds yEcor gSaA (1)

Assertion (A) : Vectors 𝑎⃗ = 2i - j +5 k and 𝑏⃗ =5i + j -3 k are perpendicular to each other .

dkj.k Reason (R) : 𝑎⃗ × 𝑏⃗ , 𝑎⃗ vkSj 𝑏⃗ nksuksa ds fy, yEcor ,d lfn”k gSA

2 (Sin3 x + Cos3 x) dk (Sin x + Cos x) ds lkis{k vodyu Kkr dhft,A (1)

Find the differential of (Sin3 x + Cos3 x) with respect to (Sin x + Cos x)

3. ;fn x = at2, y = t3 rks dk eku Kkr dhft, A (1)
d2y
If x = at2, y = t3 then find :
dx2

4 ∫ ex ( & ) dx Kkr dhft,A (1)

Find ∫ ex ( & ) dx

5 ∫ Sin -12x dx Kkr dhft,A (1)

Find ∫ Sin -12x dx

6 vody lehdj.k y' -2x-2=0 dk gy Kkr dhft,A (1)

Find the solution of differential equation y' -2x-2=0

7 ;fn nks ek=d lfn”kksa dk ;ksx ,d ek=d lfn”k gks ] rks fl) dhft, fd mu nks lfn”kksa ds
vUrj dk ifjek.k √3 gksxkA (2)
The sum of two unit vectors is a unit vector . Then prove that the megnitude of the difference of
those two vectors will be √3
8 fl) dhft, (2)

Prove that
Sin -1(2x /1+x2 )= 2 tan -1 x

9 k dk eku Kkr dhft, rkfd fuEufyf[kr Qyu x=4 ij larr gks& (2)
Find the value of k so that the following function is continuous at x=4

f(x) = , x≠4

k, x=4
10 fn[kkb, fd Qyu
y=a Cos(log x) +b Sin (log x)

d2y
vody lehdj.k 𝑥 +x +y=0 dk gy gSA (2)
dx2
Show that the function y=a Cos(log x) +b Sin (log x)

is the solution of differential equation 𝑥 +x +y=0

11 fcUnq P(1,6,3) ls js[kk = = ij Mkys x, yEc dk ikn Kkr dhft, rFkk
fcUnq P ls bldh nwjh Hkh Kkr dhft,A (2)
Find the foot of the tent pitched on the line = = from the pointP(1,6,3) . Find its distance
from the point P.

12. ,d vufHkur ikls dks nks ckj mNkyk x;kAeku ysa A ?kVuk “igyh mNky ij fo’ke la[;k
izkIr gksuk” vkSj B ?kVuk “f}rh; mNky ij fo’ke la[;k izkIr gksuk” n”kkZrs gSaA?kVukvksa A vkSj
B ds LokrU+=; dk ijh{k.k dhft,A (2)
An unbiased dice is thrown twice . Let A be the event “getting an odd number on the first throw”
and B be the event “ getting an odd number on the second throw” . Test the independence of events
A and B .

13. ;fn lEcU/k R okLrfod la[;kvksa ds leqPp; S ij bl izdkj ifjHkkf’kr gS fd]
R=ø(a,b): a2+b2=1À tgk¡ a,ab∈ S (4)

rks fn, x, laca/k dh rqY;rk dh tkap dhft,A
If relation R is defined on the set S of real numbers such that ,
R=ø(a,b): a2+b2=1À where a,ab∈ S
Then check the equivalence of given relation .
2 −4 −2
14 . vkO;wg A= 1 4 3 dks ,d lefer vkO;wg rFkk ,d fo’ke lefer vkO;wg
1 −3 2
ds ;ksxQy ds :Ik esa O;Dr dhft,A (4)
2 −4 −2
Express the matrix A= 1 4 3 as the sum of symmetric matrix and skew
1 −3 2
symmetric matrix .
15. 2 m ÅapkbZ dk vkneh 6 m Åaps fctyh ds [kacs ls nwj 5 km/h dh leku pky ls pyrk
gSA
mldh Nk;k dh yEckbZ dh o`f) nj Kkr dhft,A (4)
A man of height 2 m walks away from a 6 m high electric pole at a uniform speed of 5
Km/h. Find the rate of increase of the length of his shadow .

vFkok / Or
mu varjkyksa dks Kkr djks ftuesa Qyu
F(x)= x4 - x3 – 3x2 + x +11

(a) o/kZeku (b)gkleku gS A
Find the intervals in which the function

F(x)= 3 x4- 4 x3 – 3x2 + 36 x +11 is
10 5 5
(a) Increasing (b) Decreasing

16 . ∫ Kkr dhft,A (4)
√

Find . ∫

√
vFkok / Or
∫ ( )( )
Kkr dhft,
Find ∫ ( )( )

17. js[kk y=3x+2 , x v{k ,oa dksfV;ksa x=-1 ,oa x=1 ls f?kjs {ks=
+ dk {ks=Qy Kkr
dhft,A (4)
Find the area of the region bounded by the line y=3x+2 , the x axis and the ordinates x=-1 and x=1
.

18 . ;fn lfn”k 𝑎⃗ vkSj 𝑏⃗ rFkk 𝑐⃗ rhuksa ,d nwljs ds yEcor gksa rFkk budk ifjek.k ¼ekikad½
leku gks rks fl) dhft, fd¼ 𝑎⃗ + 𝑏⃗+ 𝑐⃗ ½ lfn'kksa 𝑎⃗, 𝑏⃗ rFkk 𝑐⃗ ls leku dks.k cukrk gSA (4)
If the vectors 𝑎⃗ , 𝑏⃗ and 𝑐⃗ are perpendicular to each other and have the same magnitude ,
then prove that ¼ 𝑎⃗ + 𝑏⃗+ 𝑐⃗ ½ makes equal angles with the vectors 𝑎⃗ , 𝑏⃗ and 𝑐⃗ .
vFkok / Or
fl) dhft, fd nks lfn'kksa 𝑎⃗ vkSj 𝑏⃗ ds fy, 𝑎⃗ + 𝑏⃗ ≤ |𝑎⃗| + 𝑏⃗
Prove that for two vectors 𝑎⃗ and 𝑏⃗ , 𝑎⃗ + 𝑏⃗ ≤ |𝑎⃗| + 𝑏⃗

19 . fl) dhft, fd nh xbZ fr;Zd ÅpkbZ vkSj egRre vk;ru okys “kadq dk v)Z “kh’kZ dks.k
tan -1√2 gksrk gSA (5)
Prove that the semi apical angle of a cone of given slant height and maximum volume is tan √2 .
-1

vFkok / Or
unh ds lh/ks fdukjs ls yxs fn, gq, {ks=Qy okys ,d vk;rkdkj [ksr ds rhu vksj rkj yxkuh
gSAunh ds fdukjs dh rjQ dksbZ rkj yxkus dh vko”;drk ughsa gSAfl) dhft, fd ckdh rhu rjQ
rkj yxkus ds fy, de ls de rkj dh vko”;drk rc gksxh] tc [ksr dh yEckbZ mldh pkSM+kbZ ls
nqxuh gksA
A rectangular field of a given area , directly adjacent to a river bank , is to be fenced on three sides,
no wire is required on the river bank . Prove that the minimum amount of wire required for the other
three sides is , when the length of the field is longer than its width .

20 . vody lehdj.k (tan -1y- x) dy = (1+y2) dx
dk fof”k’V gy Kkr dhft,A tcfd x=0 rFkk y=0 gSA (5)

Find the particular solution of the differential equation (tan -1y- x) dy = (1+y2) dx
When x=0 and y=0 .

21- js[kkvksa = = rFkk = = ds e/; U;wure nwjh ]U;wure nwjh
js[kk lehdj.k rFkk mu fcUnqvksa ds funsZ”kkad Kkr dhft, tgk U;wure nwjh js[kk nh xbZ js[kkvksa ls
feyrh gSA (5)
Find the shortest distance between the lines = = and = = ,shortest
distance line equation and coordinates of the points where the shortest distance line meets the given
line .
vFkok / Or
js[kkvksa

𝑟⃗ = 𝚤̀ + 2𝚥 + 3𝑘 + 𝜆 2𝚤̀ + 3𝚥 + 4𝑘
rFkk
𝑟⃗ = 2𝚤̀ + 4𝚥 + 5𝑘 + 𝜇 4𝚤̀ + 6𝚥 + 48
ds chp dh U;wure nwjh Kkr djksA

Find the shortest distance between the lines given by
𝑟⃗ = 𝚤̀ + 2𝚥 + 3𝑘 + 𝜆 2𝚤̀ + 3𝚥 + 4𝑘
𝑟⃗ = 2𝚤̀ + 4𝚥 + 5𝑘 + 𝜇 4𝚤̀ + 6𝚥 + 48
22. vkys[kh; fof/k ls fuEufyf[kr O;ojks/kksa ds varxZr dk vf/krdehdj.k Kkr dhft,A (5)
Determine graphically the maximum value of the given objective function subject to constraints :

mn~n”s ; Qyu / Objective function : z =1000x + 600y
O;ojks/k / Constraints : x + y ≤ 200
4x + y ≤ 0
x ≥ 20 , x ≥ 0 , y ≥ 0
23 . ,d ijh{kk esa ,d ijh{kkFkhZ pkj fodYi okys ,d cgqfodYih; iz”u dk mRrj ;k rks vuqeku
yxkrk gS];k udy djrk gS ;k tkurk gSAmlds vuqeku yxkus dh izkf;drk 1@3 gSAvkSj mlds
}kjk udy djus dh izkf;drk 1@6 gSA;g ns[krs gq, fd mlus mRrj dh udy dh gS]mlds mRRrj
ds lgh gksus dh izkf;drk 1@8 gSA bl ckr dh izkf;drk Kkr dhft, fd mlus iz”u dk lgh
mRrj fn;k gS]tc fd og iz”u dk lgh mRrj tkurk FkkA (5)
In a test , an examinee either guesses or copies or knows the answer to a multiple choice
question with four choices . The probability that he makes a guess is 1/3 and the probability that he
copies the answer is 1/6 .The probability that his answer is correct , given that he copied it , is 1/8
.Find the probability that he knew the answer to the question , given that he correctly answered it .

24. nks fo|ky; P vkSj Q vius pqus gq, Nk=ksa dks vuq”kklu ]fouezrk vkSj le;fu’Bk ds ewY;ksa ds
vk/kkj ij iqjLd`r djuk pkgrs gSAa fo|ky; P vius 3] 2 vkSj 1 N a k= dks rhu ewY;ksa ds fy,]
izR;sd ds fy, dze”k% x :i;s ] y :Ik;s vkSj z :Ik;s nsdj dqy iqjLdkj jkf”k 1000 :Ik;s nsdj
iq:Ldr djuk pkgrk gSA fo|ky; Q vius 4] 1 vkSj 3 aNk= dks dze”k% rhu ewY;ksa ds fy, 1500
:i;s ] [kpZ djds iqjLd`r djuk pkgrk gS (rhu ewY;ksa ds fy, igys dh rjg gh iqjLdkj jkf”k
nsdj)A izR;sd ewY; ij ,d iqjLdkj ds fy, dqy jkf”k : 600 gSA
mijksDr tkudkjh ds vk/kkj ij fuEufyf[kr iz”uksa ds mRrj nhft,%
(i) mijksDr fLFfr dks n”kkZus okyk vkO;wg lehdj.k D;k gS ?
(ii) x dk eku D;k gS ?
(iii) z dk eku D;k gS ?
(iv) y dk eku D;k gS ?
(v) fo|ky; Q vuq”kklu ij iqjLdkj nsus ds fy, fdruk /ku [kpZ dj jgk gS ? (5)
Two Schools P and Q want to award their selected students on the values of discipline , politeness
and punctuality . The school P wants to award Rs. x each ,Rs. y each and Rs z each. For the three
respective values to its 3,2 and 1 students with a total awarded money of Rs 1000. School Q wants
to spend Rs 1500 to award its 4,1 and 3 students on the respective values ( by giving the same award
money for three values as before ). The total amount of awards for one prize on each value is 600.
Based on the above information ,answer the following questions :

(i) What is the matrix equation representing the above situation ?
(ii) What is the value of x ?
(iii) What is the value of z ?
(iv) What is the value of y ?
(v) How much money School Q is spending to award on discipline ?

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